Answer
$f'(x)=\frac{\sec^2\sqrt x}{2\sqrt x}$
Work Step by Step
$f(x)=tan\sqrt x$
$f(x)=tan(x^{\frac{1}{2}})$
$f'(x)=sec^2(x^{\frac{1}{2}})\times\frac{d}{dx}(x^{\frac{1}{2}})$
$f'(x)=sec^2(x^{\frac{1}{2}})\times\frac{1}{2}x^{\frac{1}{2}-1}$
$f'(x)=sec^2(\sqrt x)\times\frac{1}{2}x^{\frac{-1}{2}}$
$f'(x)=sec^2\sqrt x)\times\frac{1}{2{x^\frac{1}{2}}}$
$f'(x)=sec^2(\sqrt x)\times\frac{1}{2\sqrt x}$
$f'(x)=\frac{sec^2\sqrt x}{2\sqrt x}$